On the Complete Analogy of Complex Analysis and Real Analysis in the Field of Vectors V₂

1. Introduction

A geometric algebra (Clifford algebra) is an extension of elementary algebra to work with geometrical objects such as vectors. It is built out of two fundamental operations addition and geometric product, [2]. The multiplication of vectors alone results in objects called multivectors, among which are bivectors, the name applied in this paper to the objects of the bivector field ${\mathbb{R}}^2$, corresponding to the field of vectors ${\mathbf{V}}_2$. Compared with other formalisms for manipulating geometric objects, geometric algebra supports dividing by a vector. The geometric product was first mentioned by Grassmann, who founded the so-called external algebra, [3]. After that, Clifford himself greatly expanded upon Grassmann’s work, to form geometric algebra, named after him Clifford algebra [2], by unifying both Grassmann’s algebra and Hamilton’s quaternion algebra. In the middle of the 20th century, Hestenes repopularized the term geometric algebra [4,5].

On the other hand, although rarely used explicitly, a geometric representation of complex numbers is implicitly based on its structure of the Euclidean 2-dimensional vector space. If the binary operation of the product $\overline{w}z$ of two complex numbers $\overline{w}$ and z is considered as the sum of the inner product $w\circ z=(\overline{w}z+w\overline{z})/2=\mathbf{1}Re\left(\overline{w}z\right)$ and outer product $w\wedge z=(\overline{w}z-w\overline{z})/2=$į$Im\left(\overline{w}z\right)$, where $\mathbf{1}=(1,i0)$ and į$=(0,i)$, and i is an imaginary unit, it can be said that $\overline{w}z$ is in the form of a geometric product of two ivectors (two complex numbers), as two geometric objects belonging to the ivector field $\mathbb{C}$ (to the field of complex numbers $\mathbb{C}$). For any complex number z, its absolute value $\left|z\right|$ is its Euclidean norm denoted by r, and the argument $argz$ is the polar angle $\phi$. Since ordered pairs represent both complex numbers and vectors, the binary operation of the product of two complex numbers (two ordered pairs), in the form of the geometric product $wz$ ($wz=\overline{w}\circ z+\overline{w}\wedge z$), will be the basis for modifying Grassmann’s geometric product of vectors, which is defined as the sum of the inner (scalar) and outer (vector) products of two vectors. By this modification, the geometric product of two vectors becomes commutative, similar to the product of complex numbers themselves, which still supports vector division. In this manner, a complete analogy is established between the algebra of complex numbers and the modified Clifford algebra in the Euclidean 2-dimensional field of vectors ${\mathbf{V}}_2$. On the basis of that analogy, the paper presents the most important vector integral identities, in the real Euclidean field of vectors ${\mathbf{V}}_2$, which are vector analogies to the well-known integral identities of complex analysis.

1.1. Realireal Vector Space $V_{\mathbb{C}}$

The ordered pairs $\mathbf{1}=(1,i0)$ and į$=(0,i)$ are the basis of the 2-dimensional realireal vector space $V_{\mathbb{C}}$[6], which is the Cartesian product $V_{\mathbb{R}}\times i{V}_{\mathbb{R}}$ of a 1-dimensional real vector space $V_{\mathbb{R}}$ and a 1-dimensional ireal vector space $i{V}_{\mathbb{R}}$, and as such $V_{\mathbb{C}}$ is an additive Abelian (commutative) group of elements $\left(x,iy\right)=\mathbf{1}x+$į$y$. As both of these 1-dimensional vector spaces are defined over the field of real numbers $\mathbb{R}$, the real vector space $V_{\mathbb{R}}$ can be said to be a field of real numbers $\mathbb{R}$, whereas the ireal vector space $i{V}_{\mathbb{R}}$ cannot be a field, and therefore not a field of imaginary numbers $iy$. On the other hand, if the vector space $V_{\mathbb{C}}$ is complemented by a binary operation of the product of two elements $\left(a,ib\right)$ and $\left(c,id\right)$, which corresponds to the matrix product, in such a manner that

$$\left(a,ib\right)\left(c,id\right)\fallingdotseq \left[\begin{array}{cc}a& ib\\ ib& a\end{array}\right]\left[\begin{array}{cc}c& id\\ id& c\end{array}\right]=$$

(1)

$$=\left[\begin{array}{cc}ac-bd& i(ad+bc)\\ i(ad+bc)& ac-bd\end{array}\right]\risingdotseq (ac-bd,i(ad+bc)),$$

where both the commutative axiom of multiplication and the associative axiom of multiplication are satisfied, as well as the distributive axiom, and in addition the element $(x,-iy)/(x^2+y^2)$, which corresponds to the inverse matrix ${\left[\begin{array}{cc}x& iy\\ iy& x\end{array}\right]}^{-1}$, is the inverse element of the element $(x,iy)\ne (0,i0)$, then the 2-dimensional realireal vector space $V_{\mathbb{C}}$ can be said to be defined over the field of complex numbers $\mathbb{C}$, that is, $V_{\mathbb{C}}$ is the field of complex numbers $\mathbb{C}$. More precisely, on the one hand, the ordered pair $(V_{\mathbb{C}},\mathbb{R})$ is the vector space $V_{\mathbb{C}}$ over the scalar field of real numbers $\mathbb{R}$, and on the other hand, after complementation with the binary operation of the product of the elements, the ordered pair $(V_{\mathbb{C}},\mathbb{C})$ is the vector space over the ivector field $\mathbb{C}$Accordingly, complex numbers z can be said to be ivectors $\left(x,iy\right)=\mathbf{1}x+$į$y$, the elements of the vector space $V_{\mathbb{C}}$, that is, of the ivector field $\mathbb{C}$ and as such can be multiplied either by real numbers as scalars or by complex numbers z as ivectors. When the ivector $\left(a,ib\right)$ is multiplying by the imaginary unit i, then the order of of the elements, in the resulting ordered pair, is changed, so the resulting ordered pair is not an element of the ivector field $\mathbb{C}$. Therefore, from that perspective, multiplying complex numbers by imaginary numbers is absolutely unacceptable. A complex number can be multiplied by the ivector į, so that $\left(x,iy\right)$į= į$\left(x,iy\right)=-y\mathbf{1}+x$į$=(-y,ix)$ and į² = įį$=-\mathbf{1}$.

How the operator cįs·=$\left(cos·,isin·\right)=\mathbf{1}cos·+\text{į}sin·$ has the most important properties of an exponential function, since cįs·cįs·= cįs$(·+·)$, (cįs${·)}^{-1}=$cįs$(-·)$ and d(cįs$·)$=įcįs·d·, the operator $exp($į$·)$ can be said to be the exponential form of the operator cįs·. For above reason, multiplication of the operator cįs·= exp(į·) by imaginary numbers is unacceptable, but multiplication by either a real number or an ivector is acceptable.

1.2. Algebraic Structure of the Field of Real Vectors ${\mathbf{V}}_2$

The basis of the 2-dimensional real vector space $V_{{\mathbb{R}}^2}$ (the Cartesian square of the 1-dimensional real vector space $V_{\mathbb{R}}$), as an additive Abelian group of elements $\left(x,y\right)$, consists of ordered pairs $\mathbf{1}=(1,0)$ and $\widehat{\mathbf{1}}=(0,1)$, such that $\left(x,y\right)=x\mathbf{1}+y\widehat{\mathbf{1}}$. Analogous to the binary operation of the product of the elements, which we used to complement the realireal vector space $V_{\mathbb{C}}$, it is also possible to complement the vector space $V_{{\mathbb{R}}^2}$, with the binary operation of the product of two elements $\left(a,b\right)$ and $\left(c,d\right)$, which corresponds to the matrix product, in such a manner that

$$\left(a,b\right)\left(c,d\right)\fallingdotseq \left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}c& d\\ -d& c\end{array}\right]=$$

(2)

$$=\left[\begin{array}{cc}ac-bd& ad+bc\\ -(ad+bc)& ac-bd\end{array}\right]\risingdotseq (ac-bd,ad+bc),$$

where the element $(x/(x^2+y^2),-y/(x^2+y^2))$, which corresponds to the inverse matrix ${\left[\begin{array}{cc}x& y\\ -y& x\end{array}\right]}^{-1}$, is the inverse element of the vector space $V_{{\mathbb{R}}^2}^{*}=V_{{\mathbb{R}}^2}\setminus \left\{(0,0)\right\}$. In this case both the commutative axiom of multiplication and the associative axiom of multiplication are satisfied, as well as the distributive axiom. Therefore, on the one hand, the ordered pair $(V_{{\mathbb{R}}^2},\mathbb{R})$ is the vector space $V_{{\mathbb{R}}^2}$ over the scalar field of real numbers $\mathbb{R}$, whereas on the other hand, the ordered pair $(V_{{\mathbb{R}}^2},{\mathbb{R}}^2)$ is a vector space over the bivector field ${\mathbb{R}}^2$, that is, $V_{{\mathbb{R}}^2}$ is the bivector field ${\mathbb{R}}^2$, whose elements are the bivectors $\left(x,y\right)=x\mathbf{1}+y\widehat{\mathbf{1}}$. Accordingly, the bireal numbers w are bivectors $\left(x,y\right)$, elements of the vector space $V_{{\mathbb{R}}^2}$, that is, of the bivector field ${\mathbb{R}}^2$ and as such can be multiplied either by real numbers as scalars or by bireal numbers as bivectors.

The field of vectors ${\mathbf{V}}_2$ corresponds to the bivector field ${\mathbb{R}}^2$, in the sense of the correspondence: $\mathbf{1}\fallingdotseq$$\mathbf{e}$ and $\widehat{\mathbf{1}}\fallingdotseq$$\widehat{\mathbf{e}}$, where the unit vectors $\mathbf{e}$ and $\widehat{\mathbf{e}}$ are orthogonal basis vectors of the field of vectors ${\mathbf{V}}_2$. The vectors $\mathbf{r}=x\mathbf{e}+y\widehat{\mathbf{e}}$, as elements of the field ${\mathbf{V}}_2$, correspond to the bireal numbers $w\in {\mathbb{R}}^2$. In other words, there is a one to one correspondence between the field of vectors ${\mathbf{V}}_2$ and the set of bireal numbers ${\mathbb{R}}^2$. If $\left(x,-y\right)\fallingdotseq x\mathbf{e}-y\widehat{\mathbf{e}}=\overline{\mathbf{r}}$, then $r=∥\overline{\mathbf{r}}∥=∥\left(x,-y\right)∥=\left|\sqrt[2]{x^2+y^2}\right|$ is the norm over the field of vectors ${\mathbf{V}}_2$ and over the bivector field ${\mathbb{R}}^2$, simultaneously. In addition, $\overline{\mathbf{r}}\mathbf{r}\risingdotseq \left(x,-y\right)\left(x,y\right)=r^2\mathbf{1}\fallingdotseq r^2\mathbf{e}$ and $r^2{\mathbf{r}}^{-1}=\overline{\mathbf{r}}$. It is quite clear that the inverse element ${\mathbf{r}}^{-1}$ allows division by a vector in the field of vectors ${\mathbf{V}}_2$.

The binary operation of the product of two vectors ${\overline{\mathbf{r}}}_1=a\mathbf{e}-b\widehat{\mathbf{e}}\in {\mathbf{V}}_2$ and ${\mathbf{r}}_2=c\mathbf{e}+d\widehat{\mathbf{e}}\in {\mathbf{V}}_2$, as follows

$${\overline{\mathbf{r}}}_1{\mathbf{r}}_2={\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2+{\mathbf{r}}_1{\overline{\mathbf{r}}}_2)+{\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2-{\mathbf{r}}_1{\overline{\mathbf{r}}}_2):=({\overline{\mathbf{r}}}_1·{\overline{\mathbf{r}}}_2)\mathbf{e}-({\overline{\mathbf{r}}}_1\times {\overline{\mathbf{r}}}_2)\times \mathbf{e}=$$

(3)

$$=({\mathbf{r}}_1·{\mathbf{r}}_2)\mathbf{e}+({\mathbf{r}}_1\times {\mathbf{r}}_2)\times \mathbf{e}=[(a\mathbf{e}+b\widehat{\mathbf{e}})·(c\mathbf{e}+d\widehat{\mathbf{e}})]\mathbf{e}+[(a\mathbf{e}+b\widehat{\mathbf{e}})\times (c\mathbf{e}+d\widehat{\mathbf{e}})]\times \mathbf{e},$$

where ${\mathbf{r}}_1·{\mathbf{r}}_2=$$∥{\mathbf{r}}_1∥∥{\mathbf{r}}_2∥cos\theta$ and $({\mathbf{r}}_1\times {\mathbf{r}}_2)\times \mathbf{e}=∥{\mathbf{r}}_1∥∥{\mathbf{r}}_2∥sin\theta \widehat{\mathbf{e}}$, which is obviously commutative and corresponds to the product $\left(a,-b\right)\left(c,d\right)=(ac+bd,ad-bc)$ of two bivectors $\left(a,-b\right)$ and $\left(c,d\right)$, is the geometric product of these two vectors. Here,

$${\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2+{\mathbf{r}}_1{\overline{\mathbf{r}}}_2):=({\mathbf{r}}_1·{\mathbf{r}}_2)\mathbf{e}={\mathbf{r}}_1\circ {\mathbf{r}}_2\risingdotseq \left(a,b\right)\circ \left(c,d\right)\mathrm{and}$$

(4)

$${\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2-{\mathbf{r}}_1{\overline{\mathbf{r}}}_2):=({\mathbf{r}}_1\times {\mathbf{r}}_2)\times \mathbf{e}={\mathbf{r}}_1\wedge {\mathbf{r}}_2\risingdotseq \left(a,b\right)\wedge \left(c,d\right).$$

(5)

In addition, $\left(a,-b\right)\left(c,d\right)=\left(a,b\right)\circ \left(c,d\right)+\left(a,b\right)\wedge \left(c,d\right)$ and

$${[\left(a,b\right)\circ \left(c,d\right)]}^2-{[\left(a,b\right)\wedge \left(c,d\right)]}^2=\left(a,-b\right)\left(c,d\right)\left(c,-d\right)\left(a,b\right)=$$

(6)

$$=\left(a,-b\right)\left(c,d\right)\left(a,b\right)\left(c,-d\right)={∥\left(a,b\right)∥}^2{∥\left(c,d\right)∥}^2\mathbf{1}.$$

The previously introduced concepts of the geometric product and bivector are closely related to the same concepts in Clifford algebra [2]. However, there is evidently a crucial difference between Clifford algebra and the algebra of the field of vectors ${\mathbf{V}}_2$, which is reflected in the fact that the geometric product of the elements of the field of vectors ${\mathbf{V}}_2$, on the one hand, is the element of the field of vectors ${\mathbf{V}}_2$, corresponding to the bivector, the element of the bivector field ${\mathbb{R}}^2$, and on the other hand, it is also commutative, which means that the field of vectors ${\mathbf{V}}_2$, in addition to being an additive Abelian group, is also a multiplicative Abelian group.

If ${\mathbf{r}}_0$ is the unit vector of the vector $\mathbf{r}$$(\mathbf{r}=$$r{\mathbf{r}}_0)$, then ${\mathbf{r}}_0^{-1}={\overline{\mathbf{r}}}_0$ and

$${\displaystyle \frac{{\mathbf{r}}_{0_1}}{{\mathbf{r}}_{0_2}}}={\mathbf{r}}_{0_1}{\overline{\mathbf{r}}}_{0_2}=({\mathbf{r}}_{0_1}·{\mathbf{r}}_{0_2})\mathbf{e}-({\mathbf{r}}_{0_1}\times {\mathbf{r}}_{0_2})\times \mathbf{e},$$

(7)

that is,

$${\mathbf{r}}_{0_1}\circ {\mathbf{r}}_{0_2}={\displaystyle \frac{1}{2}}({\displaystyle \frac{{\mathbf{r}}_{0_2}}{{\mathbf{r}}_{0_1}}}+{\displaystyle \frac{{\mathbf{r}}_{0_1}}{{\mathbf{r}}_{0_2}}})\mathrm{i}{\mathbf{r}}_{0_1}\wedge {\mathbf{r}}_{0_2}={\displaystyle \frac{1}{2}}({\displaystyle \frac{{\mathbf{r}}_{0_2}}{{\mathbf{r}}_{0_1}}}-{\displaystyle \frac{{\mathbf{r}}_{0_1}}{{\mathbf{r}}_{0_2}}}),$$

(8)

where ${\mathbf{r}}_{0_1}\circ {\mathbf{r}}_{0_2}=({\mathbf{r}}_{0_1}·{\mathbf{r}}_{0_2})\mathbf{e}$ i ${\mathbf{r}}_{0_1}\wedge {\mathbf{r}}_{0_2}=({\mathbf{r}}_{0_1}\times {\mathbf{r}}_{0_2})\times \mathbf{e}$ are the symmetric and antisymmetric parts of the geometric product ${\overline{\mathbf{r}}}_{0_1}{\mathbf{r}}_{0_2}$, respectively. Therefore, to emphasize once again, the geometric product of two vectors is a vector in ${\mathbf{V}}_2$. Accordingly, $\mathbf{re}=(\overline{\mathbf{r}}·\mathbf{e})\mathbf{e}+(\overline{\mathbf{r}}\times \mathbf{e})\times \mathbf{e}=x\mathbf{e}+y\widehat{\mathbf{e}}=\mathbf{r}$ and $\mathbf{r}\widehat{\mathbf{e}}=(\overline{\mathbf{r}}·\widehat{\mathbf{e}})\mathbf{e}+(\overline{\mathbf{r}}\times \widehat{\mathbf{e}})\times \mathbf{e}={\mathbf{r}}_{\perp }$. The vector ${\mathbf{r}}_{\perp }$ is orthogonal to the vector $\mathbf{r}$, that is, ${\mathbf{r}}_{\perp }$ is the vector obtained by rotating the vector $\mathbf{r}=r{\mathbf{r}}_0=r(\mathbf{e}cos\phi +\widehat{\mathbf{e}}sin\phi )$, where $\phi$ is the angle between the vector $\mathbf{r}$ and the vector $\mathbf{e}$, by $\pi /2$ radians in the positive mathematical direction. In addition, the unit vector ${\overline{\mathbf{r}}}_0=cos\phi \mathbf{e}-sin\phi \widehat{\mathbf{e}}$ is the inverse vector ${\mathbf{r}}_0^{-1}$ of the unit vector ${\mathbf{r}}_0$ of the vector $\mathbf{r}$, and therefore also the unit vector of the inverse vector ${\mathbf{r}}^{-1}$ of the vector $\mathbf{r}$, so that ${\mathbf{r}}^{-1}={\overline{\mathbf{r}}}_0/r$ and $\overline{\mathbf{r}}=r{\overline{\mathbf{r}}}_0$.

The main purpose of the paper is to derive vector integral identities, in the field of vectors ${\mathbf{V}}_2$, which are analogous to the known integral identities of complex analysis, on the basis of the analogy of the ivector field $\mathbb{C}$ and the bivector field ${\mathbb{R}}^2$, which corresponds to the field of vectors ${\mathbf{V}}_2$.

2. Fundamental Integral Identities in ${\mathbf{V}}_2$

As c$\widehat{\mathbf{e}}$s·c$\widehat{\mathbf{e}}$s·=c$\widehat{\mathbf{e}}$s$(·+·)$, (c$\widehat{\mathbf{e}}$s${·)}^{-1}=$c$\widehat{\mathbf{e}}$s$(-·)$ i d(c$\widehat{\mathbf{e}}$s$·)=\widehat{\mathbf{e}}$c$\widehat{\mathbf{e}}$s·d·, the bivector operator c$\widehat{\mathbf{e}}$s·$=\mathbf{e}cos·+\widehat{\mathbf{e}}sin·$ also has the properties of an exponential function, similar to the ivector operator cįs·. The operator $exp(\widehat{\mathbf{e}}·)$ is the exponential form of the operator c$\widehat{\mathbf{e}}$s·. Since ${\mathbf{r}}_0$=c$\widehat{\mathbf{e}}$s$\phi =exp\left(\widehat{\mathbf{e}}\phi \right)$, one more analogy with complex analysis is the notion of the so-called vector logarithmic function $log\mathbf{r}=lnr\mathbf{e}+\phi \widehat{\mathbf{e}}$, where $2\phi \widehat{\mathbf{e}}=2log{\mathbf{r}}_0=$$log({\mathbf{r}}_0/{\overline{\mathbf{r}}}_0)$. In addition, Log$\mathbf{r}$$=lnr\mathbf{e}+(\phi \pm 2\pi n)\widehat{\mathbf{e}}$, $n\in \mathbb{N}$. Let ${\widehat{\mathbf{r}}}_0=-{\mathbf{r}}_{0\perp }^{-1}=\mathbf{e}sin\phi +\widehat{\mathbf{e}}cos\phi =$s$\widehat{\mathbf{e}}$c$\phi =\widehat{\mathbf{e}}{\overline{\mathbf{r}}}_0$. The ordered pair of vectors $({\overline{\mathbf{r}}}_0,{\widehat{\mathbf{r}}}_0)$ is the inverse orthonormal basis with respect to the orthonormal basis $({\mathbf{r}}_0,{\mathbf{r}}_{0\perp })$ of the field of vectors ${\mathbf{V}}_2$. For an arbitrary vector $\%\in {\mathbf{V}}_2$, the vector $\%{\mathbf{r}}_0$ is the rotated vector $\%$, in the positive mathematical direction, by the angle $\phi$, and the vector $\%{\mathbf{r}}_{0\perp }$ by the angle $\pi /2+\phi$. The geometric products of the vector $\%$ with the inverse basis vectors ${\overline{\mathbf{r}}}_0$ and ${\widehat{\mathbf{r}}}_0$ rotate the vector $\%$ by the angles $-\phi$ and $\pi /2-\phi$, respectively, in the positive mathematical direction. On the basis of the geometric products $\mathbf{ee}=\mathbf{e}$, $\mathbf{e}\widehat{\mathbf{e}}=\widehat{\mathbf{e}}$ and $\widehat{\mathbf{e}}\widehat{\mathbf{e}}=-\mathbf{e}$

$${\mathbf{r}}_0{\mathbf{r}}_0=({\overline{\mathbf{r}}}_0·{\mathbf{r}}_0)\mathbf{e}+({\overline{\mathbf{r}}}_0\times {\mathbf{r}}_0)\times \mathbf{e}=({cos}^2\phi -{sin}^2\phi )\mathbf{e}+2(cos\phi sin\phi )\widehat{\mathbf{e}},$$

(9)

${\mathbf{r}}_0{\overline{\mathbf{r}}}_0=\mathbf{e}$ and ${\mathbf{r}}_0{\widehat{\mathbf{r}}}_0=\widehat{\mathbf{e}}$ ($\widehat{\mathbf{r}}=\overline{\mathbf{r}}\widehat{\mathbf{e}}$), all other combinations of geometric products of the basis vectors ${\mathbf{r}}_0$, ${\mathbf{r}}_{0\perp }$, ${\overline{\mathbf{r}}}_0$ and ${\widehat{\mathbf{r}}}_0$ can also be obtained.

If we introduce the differential operator $d=dr{\partial }_r+d\phi {\partial }_{\phi }$, then $d\mathbf{r}=dr{\mathbf{r}}_0+d\phi {\mathbf{r}}_{\perp }$. Hence, $d{\mathbf{r}}_{\perp }=\widehat{\mathbf{e}}d\mathbf{r}=dr{\mathbf{r}}_{0\perp }-d\phi \mathbf{r}$ and $d\widehat{\mathbf{r}}=\widehat{\mathbf{e}}d\overline{\mathbf{r}}=\widehat{\mathbf{e}}(dr{\overline{\mathbf{r}}}_0-d\phi \widehat{\mathbf{r}})=dr{\widehat{\mathbf{r}}}_0+d\phi \overline{\mathbf{r}}$. Since $2rcos\phi \mathbf{e}=\mathbf{r}+\overline{\mathbf{r}}$ and $2rsin\phi \widehat{\mathbf{e}}=\mathbf{r}-\overline{\mathbf{r}}$, the vector operators of partial derivatives are introduced, as a vector analogue of the Virtinger operators [16],

$${ð}_{\mathbf{r}}={\partial }_{\mathbf{r}}r{\partial }_r+{\partial }_{\mathbf{r}}\phi {\partial }_{\phi }={\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_0{\partial }_r-{\displaystyle \frac{{\widehat{\mathbf{r}}}_0}{r}}{\partial }_{\phi })\mathrm{and}{ð}_{\overline{\mathbf{r}}}={\overline{ð}}_{\mathbf{r}}={\displaystyle \frac{1}{2}}({\mathbf{r}}_0{\partial }_r+{\displaystyle \frac{{\mathbf{r}}_{0\perp }}{r}}{\partial }_{\phi }).$$

(10)

Here,

$${\partial }_{\mathbf{r}}r={\displaystyle \frac{1}{2r}}{\partial }_{\mathbf{r}}{r}^2={\displaystyle \frac{{\overline{\mathbf{r}}}_0}{2}}\mathrm{and}{\partial }_{\mathbf{r}}\phi ={cos}^2\phi {\partial }_{\mathbf{r}}tan\phi ={\left({\displaystyle \frac{\mathbf{r}+\overline{\mathbf{r}}}{2r}}\right)}^2{\displaystyle \frac{-2\widehat{\mathbf{e}}\overline{\mathbf{r}}}{{(\mathbf{r}+\overline{\mathbf{r}})}^2}}=-{\displaystyle \frac{{\widehat{\mathbf{r}}}_0}{2r}}.$$

(11)

It is important to emphasize that when geometric products and geometric quotients are differentiated, the same rules apply as when ordinary products and quotients are differentiated. Namely,

$$d{\displaystyle \frac{\mathbf{r}}{\overline{\mathbf{r}}}}=\left(d{\displaystyle \frac{1}{r^2}}\right){\mathbf{r}}^2+{\displaystyle \frac{1}{r^2}}d{\mathbf{r}}^2=-2({\displaystyle \frac{{\mathbf{r}}^2}{r^3}}dr-{\displaystyle \frac{1}{r^2}}\mathbf{r}d\mathbf{r})=$$

(12)

$$=-2[{\displaystyle \frac{{\mathbf{r}}^2}{2r^4}}(\overline{\mathbf{r}}d\mathbf{r}+\mathbf{r}d\overline{\mathbf{r}})-{\displaystyle \frac{1}{r^2}}\mathbf{r}d\mathbf{r}]=-2[{\displaystyle \frac{1}{2{\overline{\mathbf{r}}}^2}}(\overline{\mathbf{r}}d\mathbf{r}+\mathbf{r}d\overline{\mathbf{r}})-{\displaystyle \frac{1}{{\overline{\mathbf{r}}}^2}}\overline{\mathbf{r}}d\mathbf{r}]={\displaystyle \frac{\overline{\mathbf{r}}d\mathbf{r}-\mathbf{r}d\overline{\mathbf{r}}}{{\overline{\mathbf{r}}}^2}}.$$

The vector operator

$$\mathbf{d}={\mathbf{r}}_0(d\mathbf{r}{ð}_{\mathbf{r}}+d\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})=2{\mathbf{r}}_0(d\mathbf{r}\circ {ð}_{\overline{\mathbf{r}}})={\mathbf{r}}_0(dr{\partial }_r+d\phi {\partial }_{\phi })={\mathbf{r}}_{0}d,$$

(13)

with the symmetric part of the geometric product $2d\mathbf{r}{ð}_{\mathbf{r}}$, can be said to be a radial vector differential operator. The vector operator, with the antisymmetric part,

$${\mathbf{d}}_{\perp }={\mathbf{r}}_0(d\mathbf{r}{ð}_{\mathbf{r}}-d\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})=-2{\mathbf{r}}_0(d\mathbf{r}\wedge {ð}_{\overline{\mathbf{r}}})={\mathbf{r}}_{0\perp }(rd\phi {\partial }_r-{\displaystyle \frac{1}{r}}dr{\partial }_{\phi })={\mathbf{r}}_{0\perp }\widehat{d}$$

(14)

is a transverse vector differential operator. It is obvious that the vector operator $2{ð}_{\overline{\mathbf{r}}}$ is a gradient operator in polar coordinates. The symmetric part of the geometric product $2{ð}_{\mathbf{r}}\mathbf{F}$ is the divergence vector ($div$) of the vector field $\mathbf{F}=F{\mathbf{r}}_0+F_{\perp }{\mathbf{r}}_{0\perp }$, and the antisymmetric part is the $curl$ of the vector field $\mathbf{F}$, since

$${\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\circ {\overline{\mathbf{r}}}_0\mathbf{F}+{\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\wedge {\overline{\mathbf{r}}}_0\mathbf{F}={\mathbf{r}}_0{ð}_{\mathbf{r}}\left({\overline{\mathbf{r}}}_0\mathbf{F}\right)=-{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{2r}}\mathbf{F}+{ð}_{\mathbf{r}}\mathbf{F}\mathrm{and}$$

(15)

$${ð}_{\mathbf{r}}\mathbf{F}={\displaystyle \frac{1}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\circ \overline{\mathbf{r}}\mathbf{F})+{\displaystyle \frac{1}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\wedge \overline{\mathbf{r}}\mathbf{F}),$$

(16)

so that

$$2{ð}_{\mathbf{r}}\mathbf{F}={\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_0}{r}}[{\partial }_r\left(rF\right)+{\partial }_{\phi }{F}_{\perp }]+{\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_{0\perp }}{r}}[{\partial }_r\left(r{F}_{\perp }\right)-{\partial }_{\phi }F]=$$

(17)

$$={\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}div\mathbf{F}+curl\mathbf{F}\times {\overline{\mathbf{r}}}_0{\mathbf{r}}_0,$$

that is,

$$2{ð}_{\overline{\mathbf{r}}}\mathbf{F}={\displaystyle \frac{{\mathbf{r}}_0{\mathbf{r}}_0}{r}}(r{\partial }_{r}F-{\partial }_{\phi }{F}_{\perp })+{\displaystyle \frac{{\mathbf{r}}_0{\mathbf{r}}_{0\perp }}{r}}(r{\partial }_r{F}_{\perp }+{\partial }_{\phi }F)-{\displaystyle \frac{\mathbf{F}}{\overline{\mathbf{r}}}}=grad\mathbf{F},$$

(18)

where

$${\mathbf{r}}_0{ð}_{\mathbf{r}}\circ {\overline{\mathbf{r}}}_0\mathbf{F}+{\mathbf{r}}_0{ð}_{\mathbf{r}}\wedge {\overline{\mathbf{r}}}_0\mathbf{F}={\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\left({\overline{\mathbf{r}}}_0\mathbf{F}\right)={\overline{\mathbf{r}}}_0({\displaystyle \frac{1}{2r}}\mathbf{F}+{\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\mathbf{F}),$$

(19)

$$div\mathbf{F}={\displaystyle \frac{2}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}·\overline{\mathbf{r}}\mathbf{F})\mathrm{and}curl\mathbf{F}={\displaystyle \frac{2}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\times \overline{\mathbf{r}}\mathbf{F}).$$

On the other hand, as $\mathbf{r}=r$c$\widehat{\mathbf{e}}$s$\phi =rexp\left(\phi \widehat{\mathbf{e}}\right)$, it follows that $2\phi \widehat{\mathbf{e}}=$$log(\mathbf{r}/\overline{\mathbf{r}})$,

$$\mathbf{d}\phi ={\mathbf{r}}_{0}d\phi ={\mathbf{r}}_0({ð}_{\mathbf{r}}\phi d\mathbf{r}+{ð}_{\overline{\mathbf{r}}}\phi d\overline{\mathbf{r}})=-{\displaystyle \frac{{\mathbf{r}}_{0\perp }}{2}}dlog\left({\displaystyle \frac{\mathbf{r}}{\overline{\mathbf{r}}}}\right)\mathrm{and}$$

(20)

$$\mathbf{d}r={\mathbf{r}}_{0}dr={\mathbf{r}}_0\left({ð}_{\mathbf{r}}rd\mathbf{r}+{ð}_{\overline{\mathbf{r}}}rd\overline{\mathbf{r}}\right)={\mathbf{r}}_{0}d{\left(\mathbf{r}\overline{\mathbf{r}}\right)}^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\mathbf{r}}{2}}dlog\left(\mathbf{r}\overline{\mathbf{r}}\right).$$

(21)

Therefore,

$${\mathbf{r}}_{0\perp }drd\phi ={\displaystyle \frac{\mathbf{r}}{4}}dlog\left(\mathbf{r}\overline{\mathbf{r}}\right)dlog\left({\displaystyle \frac{\mathbf{r}}{\overline{\mathbf{r}}}}\right).$$

(22)

In addition,

$${\displaystyle \frac{\mathbf{r}}{4}}dlog\left(\mathbf{r}\overline{\mathbf{r}}\right)dlog\left({\displaystyle \frac{\mathbf{r}}{\overline{\mathbf{r}}}}\right)={\displaystyle \frac{\mathbf{r}}{4r^4}}(\overline{\mathbf{r}}d\mathbf{r}+\mathbf{r}d\overline{\mathbf{r}})(\overline{\mathbf{r}}d\mathbf{r}-\mathbf{r}d\overline{\mathbf{r}})={\displaystyle \frac{\mathbf{r}}{4r^2}}[{\left({\overline{\mathbf{r}}}_{0}d\mathbf{r}\right)}^2-{\left({\mathbf{r}}_{0}d\overline{\mathbf{r}}\right)}^2].$$

(23)

In accordance with above,

$${\displaystyle \frac{{\mathbf{r}}_0}{2}}(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})={\displaystyle \frac{{\mathbf{r}}_0}{4}}(\overline{\mathbf{d}}\mathbf{r}\overline{\mathbf{d}}\mathbf{r}-\mathbf{d}\overline{\mathbf{r}}\mathbf{d}\overline{\mathbf{r}})={\mathbf{r}}_{\perp }drd\phi ,$$

(24)

since $\overline{\mathbf{d}\overline{\mathbf{r}}}={\overline{\mathbf{r}}}_{0}d\mathbf{r}=\overline{\mathbf{d}}\mathbf{r}$. The vector identity just derived can be obtained explicitly, if we introduce the determinant of the Jacobi matrix (Jacobian) of the bijective mapping ${\mathbf{V}}_2\to {\mathbf{V}}_2$, defined by the system of vector equations $2lnr\mathbf{e}=log\left(\mathbf{r}\overline{\mathbf{r}}\right)$ and $2\phi \widehat{\mathbf{e}}=$$log(\mathbf{r}/\overline{\mathbf{r}})$, as follows

$$\mathbf{J}=\left|\begin{array}{cc}{\partial }_{r}log\left(\mathbf{r}\overline{\mathbf{r}}\right)& {\partial }_{\phi }log\left(\mathbf{r}\overline{\mathbf{r}}\right)\\ {\partial }_{r}log(\mathbf{r}/\overline{\mathbf{r}})& {\partial }_{\phi }log(\mathbf{r}/\overline{\mathbf{r}})\end{array}\right|=\left|\begin{array}{cc}2r^{-1}\mathbf{e}& \mathbf{0}\\ \mathbf{0}& 2\widehat{\mathbf{e}}\end{array}\right|={\displaystyle \frac{4}{r}}\widehat{\mathbf{e}}.$$

(25)

In this case, $4{\mathbf{r}}_{0\perp }drd\phi =\mathbf{rJ}drd\phi =\mathbf{r}dlog\left(\mathbf{r}\overline{\mathbf{r}}\right)dlog(\mathbf{r}/\overline{\mathbf{r}})$, which leads to (22). The vector $d\mathbf{S}=(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})/2=rdrd\phi {\mathbf{r}}_{0\perp }{\overline{\mathbf{r}}}_0=dS\widehat{\mathbf{e}}$, corresponding to the bivector $rdrd\phi \widehat{\mathbf{1}}$ of the bivector field ${\mathbb{R}}^2$, is the Lebesgue measure of the infinitesimal surface of the field of vectors ${\mathbf{V}}_2$. In accordance with above, the vector integral operator over the closed smooth Jordan curve $\gamma$, which is the boundary of an arbitrary region G in the vector field ${\mathbf{V}}_2$, is defined as follows

$$vt{\int }_{{\gamma }^{+}}^{⥀}{\overline{\mathbf{r}}}_0(\mathbf{d}+{\mathbf{d}}_{\perp })=2vt{\int }_{{\gamma }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=vt{\int }_{{\gamma }^{+}}^{⥀}[(d\mathbf{r}\circ grad)-(d\mathbf{r}\wedge grad)],$$

(26)

where $vt$ denotes the total value of an improper integral [9]-[14], such that

$$vt{\int }_{{\gamma }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=\underset{\epsilon \to 0^{+}}{lim}{\int }_{{\gamma }_{\epsilon }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=\underset{\epsilon \to 0^{+}}{lim}({\int }_{\partial S}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}+\sum _{i=1}^n{\int }_{\partial G_i}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}})=$$

(27)

$$=vp{\int }_{\gamma }^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}+\sum _{i=1}^{n_{\gamma }}\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial c({\mathbf{r}}_i,\epsilon )}^{\stackrel{⤺}{{\mathbf{r}}_{2_i}{\mathbf{r}}_{1_i}}}d\mathbf{r}{ð}_{\mathbf{r}}=vt{\int }_{{\gamma }^{-}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}+\sum _{i=1}^{n_{\gamma }}\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial c({\mathbf{r}}_i,\epsilon )}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}},$$

and $vp$ denotes the Cauchy principal value of an improper integral, ${\mathit{r}}_i$ are isolated points on the curve $\gamma$, surrounded by circles $c({\mathit{r}}_i,\epsilon )$ centered at the points ${\mathit{r}}_i$ and with an arbitrarily small radius $\epsilon$, which at the points ${\mathit{r}}_{1_i}$ and ${\mathit{r}}_{2_i}$ intersect the curve $\gamma$ and do not intersect each other. The set of points ${\mathit{r}}_i$, of Lebesgue measure zero, is the set of singular points, on curve $\gamma$, of a field on which the vector integral operator is applied. If also in region G, bounded by the curve $\gamma$, there are isolated singular points ${\varrho }_j$, which can be surrounded by circles $c({\%}_j,\epsilon )$, which do not intersect each other, then it is possible to form a simply connected region S, within which all singularities are, by connecting the circles $c({\%}_j,\epsilon )$, successively, the first with the second, the second with the third, etc., using parallel straight line segments $l_{j_1}$ and $l_{j_2}$, at a mutual distance ${\delta }_j\ll \epsilon$, as well as by connecting the circles $c({\mathit{r}}_i,\epsilon )$ on the curve $\gamma$ with the circles $c({\%}_j,\epsilon )$, using the parallel straight line segments $l_{i_1}$ and $l_{i_2}$, at a mutual distance ${\delta }_i\ll \epsilon$. The boundary $\partial S$ of the singular region S (blue region in Figure 1), inside region G, divides region G into n subregions $G_i$.

The vector integral operator

$$\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial S}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=2\pi \widehat{\mathbf{e}}\sum _{i=1}^{n_{\gamma }}Res\left({ð}_{\mathbf{r}},{\mathbf{r}}_i\right)+\sum _{i=1}^{n_{\gamma }}({\int }_{l_i^{+}}d\mathbf{r}{ð}_{\mathbf{r}}-{\int }_{l_i^{-}}d\mathbf{r}{ð}_{\mathbf{r}})+$$

(28)

$$+2\pi \widehat{\mathbf{e}}\sum _{j=1}^{n_G}Res({ð}_{\mathbf{r}},{\%}_j)+\sum _{j=1}^{n_G-1}({\int }_{l_j^{+}}d\mathbf{r}{ð}_{\mathbf{r}}-{\int }_{l_j^{-}}d\mathbf{r}{ð}_{\mathbf{r}}),$$

where

$$\underset{\epsilon \to 0^{+}}{lim}{\int }_{c({\mathbf{r}}_i,\epsilon )}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=2\pi \widehat{\mathbf{e}}Res\left({ð}_{\mathbf{r}},{\mathbf{r}}_i\right),\underset{\epsilon \to 0^{+}}{lim}{\int }_{c({\%}_j,\epsilon )}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=2\pi \widehat{\mathbf{e}}Res({ð}_{\mathbf{r}},{\%}_j),$$

(29)

$${\int }_{l_i^{+}}d\mathbf{r}{ð}_{\mathbf{r}}=\underset{\epsilon \to 0^{+}}{lim}{\int }_{l_{i_1}}d\mathbf{r}{ð}_{\mathbf{r}}\mathrm{and}{\int }_{l_i^{-}}d\mathbf{r}{ð}_{\mathbf{r}}=\underset{\epsilon \to 0^{+}}{lim}{\int }_{l_{i_2}}d\mathbf{r}{ð}_{\mathbf{r}},$$

is the residue operator in region G of the field of vectors ${\mathbf{V}}_2$. The vector integral operator in region G is as follows

$$2vt\int \underset{G^{+}}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}=2vp\int \underset{G}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}+\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial S}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=$$

(30)

$$=vp\int \underset{G}{\int }rdrd\phi ({\widehat{\mathbf{r}}}_0{\mathbf{r}}_{0}divgrad+{\widehat{\mathbf{r}}}_0{\mathbf{r}}_0\times curlgrad)+\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial S}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}},$$

where $vp\int \underset{G}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}=$${\sum }_{i=1}^n{lim}_{\epsilon \to 0^{+}}\int \underset{G_i}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}$, $2\widehat{\mathbf{r}}drd\phi =(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){\overline{\mathbf{r}}}_0$, $vt\int \underset{G^{+}}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}-vt\int \underset{G^{-}}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}=2\pi \widehat{\mathbf{e}}{\sum }_{i=1}^{n_{\gamma }}Res\left({ð}_{\mathbf{r}},{\mathbf{r}}_i\right)$ and

$${ð}_{\mathbf{r}\overline{\mathbf{r}}}^2={ð}_{\overline{\mathbf{r}}}{ð}_{\mathbf{r}}={\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_0}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}·\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})+{\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_0}{r^2}}\times (\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\times \overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})=$$

(31)

$$={\displaystyle \frac{{\overline{\mathbf{r}}}_0}{4r^2}}[{\mathbf{r}}_0[r{\partial }_r\left(r{\partial }_r\right)+{\partial }_{{\phi }^2}^2]+{\mathbf{r}}_{\perp }({\partial }_{r\phi }^2-{\partial }_{\phi r}^2)]={\displaystyle \frac{{\overline{\mathbf{r}}}_0}{4}}({\mathbf{r}}_{0}divgrad+{\mathbf{r}}_0\times curlgrad).$$

Here,

$$divgrad={\displaystyle \frac{4}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}·\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})={\displaystyle \frac{1}{r^2}}[r{\partial }_r\left(r{\partial }_r\right)+{\partial }_{{\phi }^2}^2]\mathrm{and}$$

(32)

$$curlgrad={\displaystyle \frac{4}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\times \overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})={\displaystyle \frac{{\mathbf{r}}_0\times {\mathbf{r}}_{\perp }}{r^2}}({\partial }_{\phi r}^2-{\partial }_{r\phi }^2)=\mathbf{0}.$$

(33)

If the set of singular points, either on the contour $\gamma$ or in the region G, is an empty set, the choice of a representative point ($\mathit{r}$ on the contour $\%$ or $\varrho$ in the region G, respectively) is arbitrary. If the field is uniform [7], then ${\int }_{l_i^{+}}d\mathbf{r}{ð}_{\mathbf{r}}={\int }_{l_i^{-}}d\mathbf{r}{ð}_{\mathbf{r}}$, so that the choice of representative points is not necessary.

Finally, on the basis of the result of the Kelvin-Stokes theorem (Green’s theorem) [6],

$$vt{\int }_{{\gamma }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=vt{\int }_{{\gamma }^{+}}^{⥀}[(d\mathbf{r}\circ {ð}_{\overline{\mathbf{r}}})-(d\mathbf{r}\wedge {ð}_{\overline{\mathbf{r}}})]=$$

(34)

$$={\displaystyle \frac{1}{2}}vt\int \underset{G^{+}}{\int }\widehat{\mathbf{r}}drd\phi ({\mathbf{r}}_{0}divgrad+{\mathbf{r}}_0\times curlgrad)=vt\int \underset{G^{+}}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^2.$$

If there is a limit ${lim}_{\epsilon \to 0^{+}}{\int }_{{\gamma }_{\epsilon }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}$, and ${\int }_{\partial S}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}$ tends to infinity as $\epsilon$ tends to zero, then the limit ${lim}_{\epsilon \to 0^{+}}{\sum }_{i=1}^n{\int }_{\partial G_i}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}=vp\int \underset{G}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}){ð}_{\mathbf{r}\overline{\mathbf{r}}}^{\mathbf{2}}$ is also infinite. In this emphasized case, the limit ${lim}_{\epsilon \to 0^{+}}{\int }_{{\gamma }_{\epsilon }^{+}}^{⥀}d\mathbf{r}{ð}_{\mathbf{r}}$ leads to the indeterminate form of the difference of two infinities, which has a finite value. According to (34), since

$$(\overline{\mathbf{d}}\mathbf{r}\wedge \mathbf{d}\overline{\mathbf{r}})({ð}_{\mathbf{r}\overline{\mathbf{r}}}^2-{ð}_{\overline{\mathbf{r}}\mathbf{r}}^2)=-{\widehat{\mathbf{r}}}_0{\mathbf{r}}_{0\perp }rdrd\phi ∥curlgrad∥={\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}drd\phi ({\partial }_{r\phi }^2-{\partial }_{\phi r}^2),$$

(35)

it follows that

$${\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}vt{\int }_{{\gamma }^{+}}^{⥀}d=vt{\int }_{{\gamma }^{+}}^{⥀}(d\mathbf{r}{ð}_{\mathbf{r}}+d\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})=vt\int \underset{G^{+}}{\int }(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})({ð}_{\mathbf{r}\overline{\mathbf{r}}}^2-{ð}_{\overline{\mathbf{r}}\mathbf{r}}^2)=$$

(36)

$$={\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}vt\int \underset{G^{+}}{\int }drd\phi ({\partial }_{\phi r}^2-{\partial }_{r\phi }^2).$$

If $r{\partial }_r\left(r{\partial }_r\right)=-{\partial }_{{\phi }^2}^2$, then ${ð}_{\mathbf{r}\overline{\mathbf{r}}}^2={ð}_{\overline{\mathbf{r}}\mathbf{r}}^2=\mathbf{0}$. In addition,

$${ð}_{{\mathbf{r}}^2}^2={ð}_{\mathbf{r}}{ð}_{\mathbf{r}}={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{r}}[\overline{\mathbf{r}}{\partial }_{r^2}^2-{\widehat{\mathbf{r}}}_0({\partial }_{\phi r}^2-{\displaystyle \frac{1}{r}}{\partial }_{\phi })]={\overline{\mathbf{r}}}_0{\partial }_r\left({ð}_{\mathbf{r}}\right),$$

(37)

since

$${ð}_{{\mathbf{r}}^2}^2={ð}_{\mathbf{r}}{ð}_{\mathbf{r}}={\displaystyle \frac{1}{4}}({\overline{\mathbf{r}}}_0{\partial }_r-{\widehat{\mathbf{r}}}_0{\displaystyle \frac{1}{r}}{\partial }_{\phi })({\overline{\mathbf{r}}}_0{\partial }_r-{\widehat{\mathbf{r}}}_0{\displaystyle \frac{1}{r}}{\partial }_{\phi })=$$

(38)

$$={\displaystyle \frac{{\overline{\mathbf{r}}}_0}{4r^2}}[{\overline{\mathbf{r}}}_0[r{\partial }_r\left(r{\partial }_r\right)-{\partial }_{{\phi }^2}^2-2r{\partial }_r]-\widehat{\mathbf{r}}({\partial }_{\phi r}^2+{\partial }_{r\phi }^2-{\displaystyle \frac{2}{r}}{\partial }_{\phi })].$$

Clearly, ${ð}_{{\overline{\mathbf{r}}}^2}^2={\overline{ð}}_{{\mathbf{r}}^2}^2$.

2.1. Integrals of Scalar and Vector Fields

The vector differential of a scalar field $F:{\mathbf{V}}_2\to \mathbb{R}$ is as follows

$$\mathbf{d}F={\mathbf{r}}_0({ð}_{\mathbf{r}}Fd\mathbf{r}+{ð}_{\overline{\mathbf{r}}}Fd\overline{\mathbf{r}})={\mathbf{r}}_0(\mathit{f}d\mathbf{r}+\overline{\mathit{f}}d\overline{\mathbf{r}}),$$

(39)

where $\mathit{f}={ð}_{\mathbf{r}}F$. The second vector partial derivative of F is the first vector partial derivative of the vector field $\mathit{f}$, so that

$${ð}_{\mathbf{r}}\mathit{f}={ð}_{{\mathbf{r}}^2}^{2}F={\displaystyle \frac{{\overline{\mathbf{r}}}_0}{4r^2}}[[r{\partial }_r(r{\partial }_{r}F)-{\partial }_{{\phi }^2}^{2}F-2r{\partial }_{r}F]){\overline{\mathbf{r}}}_0-$$

(40)

$$-[({\partial }_{\phi r}^{2}F+{\partial }_{r\phi }^{2}F)-{\displaystyle \frac{2}{r}}{\partial }_{\phi }F]\widehat{\mathbf{r}}]\mathrm{and}$$

$${ð}_{\overline{\mathbf{r}}}\mathit{f}={ð}_{\mathbf{r}\overline{\mathbf{r}}}^{2}F={\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_0}{4r^2}}[r{\partial }_r\left(r{\partial }_{r}F\right)+{\partial }_{{\phi }^2}^{2}F].$$

(41)

If $\mathit{f}={ð}_{\mathbf{r}}F$ and $\overline{\mathit{f}}={ð}_{\overline{\mathbf{r}}}F$ are uniform vector fields, then by applying the vector integral operator (36) to the scalar field F, a vector integral identity is obtained

$${\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}vt{\int }_{{\gamma }^{+}}^{⥀}dF=vt{\int }_{{\gamma }^{+}}^{⥀}{ð}_{\mathbf{r}}Fd\mathbf{r}+{ð}_{\overline{\mathbf{r}}}Fd\overline{\mathbf{r}}=vt{\int }_{{\gamma }^{+}}^{⥀}\mathit{f}d\mathbf{r}+\overline{\mathit{f}}d\overline{\mathbf{r}}=$$

(42)

$$=vt\int \underset{G^{+}}{\int }({ð}_{\mathbf{r}\overline{\mathbf{r}}}^{2}F-{ð}_{\overline{\mathbf{r}}\mathbf{r}}^{2}F)(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})={\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}vt\int \underset{G^{+}}{\int }({\partial }_{\phi r}^{2}F-{\partial }_{r\phi }^{2}F)drd\phi =$$

$$=\left(2\pi \widehat{\mathbf{e}}\right)[\sum _{i=1}^{n_{\gamma }}Res\left(\mathit{f},{\mathbf{r}}_i\right)+\sum _{j=1}^{n_G}Res(\mathit{f},{\%}_j)+\sum _{i=1}^{n_{\gamma }}\overline{Res}\left(\overline{\mathit{f}},{\mathbf{r}}_i\right)+\sum _{j=1}^{n_G}\overline{Res}(\overline{\mathit{f}},{\%}_j)],$$

where ${lim}_{\epsilon \to 0^{+}}{\int }_{\partial S}^{⥀}\overline{\mathit{f}}d\overline{\mathbf{r}}=\left(2\pi \widehat{\mathbf{e}}\right)[{\sum }_{i=1}^{n_{\gamma }}\overline{Res}\left(\overline{\mathit{f}},{\mathit{r}}_i\right)+{\sum }_{j=1}^{n_G}\overline{Res}(\overline{\mathit{f}},{\%}_j)]$. The integral identity of complex analysis, which is an analogue of the vector integral identity (42), is the integral identity of Cauchy’s integral theorem [7].

As ${\partial }_{\phi r}^{2}F={\partial }_{r\phi }^{2}F$ and if, in addition, $r{\partial }_r\left(r{\partial }_{r}F\right)=-{\partial }_{{\phi }^2}^{2}F$, then ${ð}_{\mathbf{r}}\overline{\mathit{f}}={ð}_{\overline{\mathbf{r}}}\mathit{f}=\mathbf{0}$, that is,

$$\mathbf{r}{\partial }_r\mathit{f}=-{\mathbf{r}}_{0\perp }{\partial }_{\phi }\mathit{f}\mathrm{and}$$

(43)

$$\mathbf{r}{ð}_{\mathbf{r}}\mathit{f}=\mathbf{r}{ð}_{{\mathbf{r}}^2}^{2}F=r{\partial }_r\left({ð}_{\mathbf{r}}F\right)=r{\partial }_r\mathit{f}.$$

(44)

A vector field $\mathit{f}$, satisfying theCauchy-Riemann condition $\mathbf{r}{\partial }_r\mathit{f}=-{\mathbf{r}}_{0\perp }{\partial }_{\phi }\mathit{f}$, is said, analogous to complex analytic functions, to be an analytic vector field. Hence, an analytic vector field is a vector derivative of the Laplace scalar field F. Clearly, the coordinate components of the analytic vector field $\mathit{f}$ are also Laplace scalar fields.

Assume that the analytic vector field $\mathit{f}$, as the vector derivative of the Laplace scalar field F, is not defined at the point $\%\in int.G$, where G is a region in the field of vectors ${\mathbf{V}}_2$, bounded by a closed smooth Jordan curve $\gamma$, as well as at point $\mathbf{r}$ on curve $\gamma$. The vector integral identity

$${\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}vt{\int }_{{\gamma }^{+}}^{⥀}dF=vt{\int }_{{\gamma }^{+}}^{⥀}{ð}_{\mathbf{r}}Fd\mathbf{r}=vt{\int }_{{\gamma }^{+}}^{⥀}\mathit{f}d\mathbf{r}=$$

(45)

$$=\left(2\pi \widehat{\mathbf{e}}\right)[Res\left(\mathit{f},\mathbf{r}\right)+Res\left(\mathit{f},\%\right)]+{\int }_{l^{+}}\mathit{f}d\mathbf{r}-{\int }_{l^{-}}\mathit{f}d\mathbf{r},$$

is a vector analogue of the integral identity of Cauchy’s integral theorem, which is slightly generalized, since in this emphasized case

$$vt\int \underset{G^{+}}{\int }{ð}_{\mathbf{r}\overline{\mathbf{r}}}^2\mathit{f}(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})=\underset{\epsilon \to 0^{+}}{lim}{\int }_{\partial S}^{⥀}\mathit{f}d\mathbf{r}=$$

(46)

$$=\left(2\pi \widehat{\mathbf{e}}\right)[Res\left(\mathit{f},\mathbf{r}\right)+Res\left(\mathit{f},\%\right)]+{\int }_{l^{+}}\mathit{f}d\mathbf{r}-{\int }_{l^{-}}\mathit{f}d\mathbf{r}.$$

If $\mathit{f}$ is a differentiable (regular) vector field, but not an analytic vector field, in an arbitrary region G of the vector field ${\mathbf{V}}_2$, bounded by a closed smooth Jordan curve $\gamma$, the integral identity

$$\underset{G\to {\%}_k}{lim}{\displaystyle \frac{1}{{\mathbf{S}}_{\gamma }}}{\int }_{\gamma }^{⥀}\mathit{f}d\mathbf{r}=2{ð}_{\overline{\mathbf{r}}}\mathit{f}\left({\%}_k\right)=grad\mathit{f}\left({\%}_k\right)$$

(47)

where $2{\mathbf{S}}_{\gamma }={\int }_{\gamma }^{⥀}\mathbf{r}\wedge d\mathbf{r}$, is a vector analogue, in the field of vectors ${\mathbf{V}}_2$, of the surface (spatial) derivative, which was introduced, into complex analysis, by Pompeiu[8], originally calling it the areolar derivative. Similarly, based on the vector identity (17), the so-called cumulative surface (spatial) derivative of the vector field $\mathit{f}$ can be defined as follows

$$\underset{G\to {\%}_k}{lim}{\displaystyle \frac{1}{{\mathbf{S}}_{\gamma }}}{\int }_{\gamma }^{⥀}\mathit{f}d\overline{\mathbf{r}}=-2{ð}_{\mathbf{r}}\mathit{f}\left({\%}_k\right)={\overline{\mathbf{r}}}_0{\mathbf{r}}_0\times curl\mathit{f}\left({\%}_k\right)-{\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}div\mathit{f}\left({\%}_k\right).$$

(48)

According to (47), if $\mathit{f}$ is a regular and uniform vector field in the $\epsilon$-neighborhood $C_{\epsilon }^0\setminus \left\{\mathbf{0}\right\}$ of its singular point $\%=\mathbf{0}$ and ${lim}_{\mathbf{r}\to \mathbf{0}}{ð}_{\overline{\mathbf{r}}}\left(r^2\mathit{f}\right)$$={\%}_0\in {\mathbf{V}}_2$, then

$$Res\left(\mathit{f},\mathbf{0}\right)={\left(2\pi \widehat{\mathbf{e}}\right)}^{-1}\underset{\epsilon \to 0^{+}}{lim}{\int }_{c_{\epsilon }^0}^{⥀}\mathit{f}d\mathbf{r}=-\underset{\epsilon \to 0^{+}}{lim}{\displaystyle \frac{\widehat{\mathbf{e}}}{2\pi {\epsilon }^2}}{\int }_{c_{\epsilon }^0}^{⥀}{r}^2\mathit{f}d\mathbf{r}=$$

(49)

$$=\underset{\epsilon \to 0^{+}}{lim}{\displaystyle \frac{1}{{\mathbf{S}}_{c_{\epsilon }^0}}}vt\int \underset{C_{\epsilon }^0}{\int }{ð}_{\overline{\mathbf{r}}}\left(r^2\mathit{f}\right)d\mathbf{S}=\underset{\mathbf{r}\to \mathbf{0}}{lim}{ð}_{\overline{\mathbf{r}}}\left(r^2\mathit{f}\right)={\%}_0.$$

If ${ð}_{\overline{\mathbf{r}}}\mathit{f}=\mathbf{0}$, then $Res\left(\mathit{f},\mathbf{0}\right)={lim}_{\mathbf{r}\to \mathbf{0}}\mathbf{r}\mathit{f}={\%}_0$, which is another vector analogy to the well-known result of complex analysis. Let $\mathit{F}$ be an analytic vector field, such that ${lim}_{r\to 0^{+}}(\mathit{F}/lnr)$ leads to the determinate form only after the application of L’Hospital’s rule n times. Then, the vector formula for $Res\left(\mathit{f},\mathbf{0}\right)$, being analogous to the complex analysis formula, can be obtained via the vector identity $\mathbf{r}\mathit{f}=r{\partial }_r\mathit{F}$, see (44), where $\mathit{f}={ð}_{\mathbf{r}}\mathit{F}$. Namely, since the same vector identity applies to the analytic vector field ${\mathbf{r}}^n\mathit{f}=r^n{\partial }_r\mathit{F}\underset{n-1}{\underbrace{{\mathbf{r}}_0...{\mathbf{r}}_0}}$, it follows that

$${ð}_{\mathbf{r}}\left({\mathbf{r}}^n\mathit{f}\right)=(n{r}^{n-1}{\partial }_r\mathit{F}+r^n{\partial }_{r^2}^2\mathit{F})\underset{n-2}{\underbrace{{\mathbf{r}}_0...{\mathbf{r}}_0}}\mathrm{and}$$

(50)

$${ð}_{{\mathbf{r}}^2}^2\left({\mathbf{r}}^n\mathit{f}\right)=[n(n-1)r^{n-2}{\partial }_r\mathit{F}+2n{r}^{n-1}{\partial }_{r^2}^2\mathit{F}+r^n{\partial }_{r^3}^3\mathit{F}]\underset{n-3}{\underbrace{{\mathbf{r}}_0...{\mathbf{r}}_0}}.$$

(51)

Accordingly, applying L’Hospital’s rule,

$${\displaystyle \frac{1}{(n-1)!}}\underset{\mathbf{r}\to \mathbf{0}}{lim}{ð}_{{\mathbf{r}}^{n-1}}^{n-1}\left({\mathbf{r}}^n\mathit{f}\right)={\displaystyle \frac{n!}{(n-1)!}}\underset{r\to 0^{+}}{lim}\sum _{k=1}^n\left(\begin{array}{c}n-1\\ n-k\end{array}\right){\displaystyle \frac{r^k{\partial }_{r^k}^k\mathit{F}}{k!}}=$$

(52)

$$=[\sum _{k=1}^n{(-1)}^{n-k}\left(\begin{array}{c}n\\ n-k\end{array}\right)]{\displaystyle \frac{{lim}_{r\to 0^{+}}{r}^n{\partial }_{r^n}^n\mathit{F}}{(n-1)!}}={\displaystyle \frac{{(-1)}^{n-1}}{(n-1)!}}\underset{r\to 0^{+}}{lim}{r}^n{\partial }_{r^n}^n\mathit{F}.$$

Further, since the vector field $\mathit{f}={\overline{\mathbf{r}}}_0{\partial }_r\mathit{F}$ is an analytic vector field, it follows that

$${ð}_{{\mathbf{r}}^{n-1}}^{n-1}\mathit{f}={\left({\displaystyle \frac{r}{\mathbf{r}}}\right)}^n{\partial }_{r^n}^n\mathit{F}\mathrm{and}$$

(53)

$$-{\displaystyle \frac{1}{(n-1)!}}\underset{\mathbf{r}\to \mathbf{0}}{lim}{(-\mathbf{r})}^n{ð}_{{\mathbf{r}}^{n-1}}^{n-1}\mathit{f}={\displaystyle \frac{{(-1)}^{n-1}}{(n-1)!}}\underset{r\to 0^{+}}{lim}{r}^n{\partial }_{r^n}^n\mathit{F}=\underset{\mathbf{r}\to \mathbf{0}}{lim}\mathbf{r}\mathit{f}.$$

(54)

This means that L’Hospital’s rule can be explicitly applied to the vector field $\mathbf{r}\mathit{f}$.

If some analytic vector field $\mathit{f}$ is regular in an arbitrary region G bounded by a closed smooth Jordan curve $\gamma$, then for the vector field

$$\mathit{f}\left(\mathbf{r}+{\%}_A\right)-[\mathit{f}\left({\%}_A\right)+\sum _{k=1}^{n-1}{ð}_{{\mathbf{r}}^k}^k\mathit{f}\left({\%}_A\right){\displaystyle \frac{{\mathbf{r}}^k}{k!}}],$$

where $\{\mathbf{0},{\%}_A\}\subset int.G$, according to (45), (52) and (54), the following is true

$$\underset{\epsilon \to 0^{+}}{lim}vt\stackrel{⥀}{{\int }_{c_{\epsilon }^0}}\{\mathit{f}\left(\mathbf{r}+{\%}_A\right)-[\mathit{f}\left({\%}_A\right)+\sum _{k=1}^{n-1}{ð}_{{\mathbf{r}}^k}^k\mathit{f}\left({\%}_A\right){\displaystyle \frac{{\mathbf{r}}^k}{k!}}]\}{\displaystyle \frac{d\mathbf{r}}{{\mathbf{r}}^{n+1}}}=$$

(55)

$$=2\pi \widehat{\mathbf{e}}\underset{\mathbf{r}\to \mathbf{0}}{lim}{\displaystyle \frac{1}{{\mathbf{r}}^n}}\{\mathit{f}\left(\mathbf{r}+{\%}_A\right)-[\mathit{f}\left({\%}_A\right)+\sum _{k=1}^{n-1}{ð}_{{\mathbf{r}}^k}^k\mathit{f}\left({\%}_A\right){\displaystyle \frac{{\mathbf{r}}^k}{k!}}]\}=2\pi \widehat{\mathbf{e}}{\displaystyle \frac{{ð}_{{\mathbf{r}}^n}^n\mathit{f}\left({\%}_A\right)}{n!}}.$$

Hence

$${\displaystyle \frac{n!}{2\pi \widehat{\mathbf{e}}}}vt\stackrel{⥀}{{\int }_{\gamma }}{\displaystyle \frac{\mathit{f}\left(\mathbf{r}+{\%}_A\right)}{{\mathbf{r}}^{n+1}}}d\mathbf{r}={ð}_{{\mathbf{r}}^n}^n\mathit{f}\left({\%}_A\right),$$

(56)

since $vt\stackrel{⥀}{{\int }_{\gamma }}{\mathbf{r}}^{-n}d\mathbf{r}=\mathbf{0}$, whenever $n\ge 2$. This is the vector analogue of the well-known Cauchy’s integral formula.

If the vector field $\mathbf{F}=F{\mathbf{r}}_0+F_{\perp }{\mathbf{r}}_{0\perp }$ is such that the scalar fields F and $F_{\perp }$ have continuous first partial derivatives in region G, bounded by the closed smooth Jordan curve $\gamma$, almost everywhere (everywhere except on the singular set of points of Lebesgue measure zero), then by applying the vector integral operator (36) to the vector field $\mathbf{F}$, one comes to the following vector integral identity

$$vt{\int }_{{\gamma }^{+}}^{⥀}d\mathbf{F}=vt{\int }_{{\gamma }^{+}}^{⥀}({ð}_{\mathbf{r}}\mathbf{F}d\mathbf{r}+{ð}_{\overline{\mathbf{r}}}\mathbf{F}d\overline{\mathbf{r}})=vt\int \underset{G^{+}}{\int }({ð}_{\mathbf{r}\overline{\mathbf{r}}}^2\mathbf{F}-{ð}_{\overline{\mathbf{r}}\mathbf{r}}^2\mathbf{F})(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})=$$

(57)

$$=\left(2\pi \widehat{\mathbf{e}}\right)[\sum _{i=1}^{n_{\gamma }}Res\left({ð}_{\mathbf{r}}\mathbf{F},{\mathbf{r}}_i\right)+\sum _{j=1}^{n_G}Res({ð}_{\mathbf{r}}\mathbf{F},{\%}_j)+\sum _{i=1}^{n_{\gamma }}\overline{Res}\left({ð}_{\overline{\mathbf{r}}}\mathbf{F},{\mathbf{r}}_i\right)+\sum _{j=1}^{n_G}\overline{Res}({ð}_{\overline{\mathbf{r}}}\mathbf{F},{\%}_j)],$$

since

$${\partial }_{\phi r}^2\mathbf{F}-{\partial }_{r\phi }^2\mathbf{F}={\mathbf{r}}_0({\partial }_{\phi r}^{2}F-{\partial }_{r\phi }^{2}F)+{\mathbf{r}}_{0\perp }({\partial }_{\phi r}^2{F}_{\perp }-{\partial }_{r\phi }^2{F}_{\perp })=\mathbf{0}\mathrm{and}$$

(58)

$${ð}_{\overline{\mathbf{r}}\mathbf{r}}^2\mathbf{F}={ð}_{\mathbf{r}\overline{\mathbf{r}}}^2\mathbf{F}={\displaystyle \frac{1}{2}}{ð}_{\overline{\mathbf{r}}}\left({\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}div\mathbf{F}+curl\mathbf{F}\times {\overline{\mathbf{r}}}_0{\mathbf{r}}_0\right)=$$

(59)

$$={\displaystyle \frac{r{\partial }_r\left(r{\partial }_r\mathbf{F}\right)+{\partial }_{{\phi }^2}^2\mathbf{F}}{4r^2}}={\displaystyle \frac{divgrad\mathbf{F}}{4}}.$$

Clearly, in the general case $curlgrad\mathbf{F}=4(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\times \overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})\mathbf{F}/r^2=\mathbf{0}$ is not the same as $curl\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)=$$4(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\times \overline{\mathbf{r}}{\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\mathbf{F})/r^2$. Namely,

$${\displaystyle \frac{{\mathbf{r}}_0}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\circ \overline{\mathbf{r}}{\overline{\mathbf{r}}}_{0}grad\mathbf{F})={\displaystyle \frac{{\mathbf{r}}_0}{2}}[{\partial }_{r^2}^{2}F+{\displaystyle \frac{1}{r^2}}({\partial }_{{\phi }^2}^{2}F-{\partial }_{\phi }{F}_{\perp })]\mathrm{and}$$

(60)

$${\displaystyle \frac{{\mathbf{r}}_0}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\wedge \overline{\mathbf{r}}{\overline{\mathbf{r}}}_{0}grad\mathbf{F})={\displaystyle \frac{{\mathbf{r}}_{0\perp }}{2}}[{\partial }_{r^2}^2{F}_{\perp }+{\displaystyle \frac{1}{r^2}}({\partial }_{{\phi }^2}^2{F}_{\perp }+{\partial }_{\phi }F)].$$

(61)

So, $divgrad\mathbf{F}=4(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}·\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}})\mathbf{F}/r^2$ differs from $div\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)=$$4(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}·\overline{\mathbf{r}}{\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\mathbf{F})/r^2$. Accordingly,

$$4{ð}_{\overline{\mathbf{r}}\mathbf{r}}^2\mathbf{F}=2{ð}_{\mathbf{r}}grad\mathbf{F}=divgrad\mathbf{F}={\displaystyle \frac{2{\mathbf{r}}_0}{r^2}}(\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\circ \overline{\mathbf{r}}{\overline{\mathbf{r}}}_{0}grad\mathbf{F}+\overline{\mathbf{r}}{ð}_{\overline{\mathbf{r}}}\wedge \overline{\mathbf{r}}{\overline{\mathbf{r}}}_{0}grad\mathbf{F})+$$

(62)

$$+{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{r}}grad\mathbf{F}={\mathbf{r}}_{0}div\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)+curl\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)\times {\mathbf{r}}_0+{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{r}}grad\mathbf{F},$$

since

$${\mathbf{r}}_0({\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\circ {\overline{\mathbf{r}}}_0{\overline{\mathbf{r}}}_{0}grad\mathbf{F}+{\overline{\mathbf{r}}}_0{ð}_{\overline{\mathbf{r}}}\wedge {\overline{\mathbf{r}}}_0{\overline{\mathbf{r}}}_{0}grad\mathbf{F})={\mathbf{r}}_0\left[{\mathbf{r}}_0{ð}_{\mathbf{r}}\left({\overline{\mathbf{r}}}_0{\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)\right]=$$

(63)

$$=-{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{r}}grad\mathbf{F}+{ð}_{\mathbf{r}}grad\mathbf{F}\mathrm{and}$$

$$2{ð}_{\mathbf{r}}\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)={\overline{\mathbf{r}}}_0{\mathbf{r}}_{0}div\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)+curl\left({\overline{\mathbf{r}}}_{0}grad\mathbf{F}\right)\times {\overline{\mathbf{r}}}_0{\mathbf{r}}_0,$$

(64)

which can be explicitly obtained if in (17) $\mathbf{F}$ is formally replaced by ${\overline{\mathbf{r}}}_{0}grad\mathbf{F}$. Therefore, the two identities 5. and 6., on page 85., in Section 3.16., Chapter 3., in [15], should be replaced by: 5. $curlgrad\mathbf{F}\equiv$$\mathbf{0}$ and 6. $divgrad\mathbf{F}=$$\mathbf{0}$ if $\mathbf{F}$ is either an analytic vector field (${ð}_{\overline{\mathbf{r}}}\mathbf{F}=\mathbf{0}$) or a Laplace vector field (${ð}_{\mathbf{r}}\mathbf{F}=\mathbf{0}$). In both of these cases, the vector field $\mathbf{F}$ satisfies Laplace’s equation $r{\partial }_r\left(r{\partial }_r\mathbf{F}\right)=-{\partial }_{{\phi }^2}^2\mathbf{F}$.

Consequently,

$$vt{\int }_{{\gamma }^{+}}^{⥀}{\displaystyle \frac{{\overline{\mathbf{r}}}_0}{2}}(\mathbf{dF}+{\mathbf{d}}_{\perp }\mathbf{F})=vt{\int }_{{\gamma }^{+}}^{⥀}{ð}_{\mathbf{r}}\mathbf{F}d\mathbf{r}=$$

(65)

$$=vt\int \underset{G^{+}}{\int }{ð}_{\mathbf{r}\overline{\mathbf{r}}}^2\mathbf{F}(\mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r})={\displaystyle \frac{{\overline{\mathbf{r}}}_0{\mathbf{r}}_{0\perp }}{2}}vt\int \underset{G^{+}}{\int }rdivgrad\mathbf{F}drd\phi .$$

On the other hand, let $\mathbf{F}=F{\mathbf{r}}_0+F_{\perp }{\mathbf{r}}_{0\perp }$ be continuous in an arbitrary region G bounded by a closed smooth Jordan curve $\gamma$, in which the partial derivatives ${\partial }_{r}F$, ${\partial }_{\phi }F$, ${\partial }_r{F}_{\perp }$and ${\partial }_{\phi }{F}_{\perp }$ exist and satisfy the Cauchy-Riemann equations

$${\partial }_{r}F={\displaystyle \frac{1}{r}}{\partial }_{\phi }{F}_{\perp }\mathrm{and}{\partial }_r{F}_{\perp }=-{\displaystyle \frac{1}{r}}{\partial }_{\phi }F.$$

(66)

Then, according to the Looman-Menchoff theorem [1], both the analytic vector field ${\overline{\mathbf{r}}}_0\mathbf{F}$ and the Laplace vector field ${\mathbf{r}}_0\overline{\mathbf{F}}$ can be said to be regular (holomorphic) vector fields in G. Therefore, on the basis of (56),

$${\left(2\pi \widehat{\mathbf{e}}\right)}^{-1}{\int }_{\gamma }^{⥀}{\displaystyle \frac{{\overline{\mathbf{r}}}_0\mathbf{F}}{{\mathbf{r}}^{n+1}}}d\mathbf{r}=Res({\displaystyle \frac{{\overline{\mathbf{r}}}_0\mathbf{F}}{{\mathbf{r}}^{n+1}}},\mathbf{0})={\displaystyle \frac{{lim}_{\mathbf{r}\to \mathbf{0}}{ð}_{{\mathbf{r}}^n}^n\left({\overline{\mathbf{r}}}_0\mathbf{F}\right)}{n!}}.$$

(67)

In addition,

$${\left(2\pi \widehat{\mathbf{e}}\right)}^{-1}[{\int }_{\gamma }^{⥀}{\displaystyle \frac{\mathbf{F}}{\mathbf{r}}}d\mathbf{r}-vp\int \underset{G}{\int }{\displaystyle \frac{{\overline{\mathbf{r}}}_{0}grad\mathbf{F}}{r}}d\mathbf{S}]=\mathbf{F}\left(\mathbf{0}\right)\mathrm{and}$$

(68)

$${\left(2\pi \widehat{\mathbf{e}}\right)}^{-1}[{\int }_{\gamma }^{⥀}{\displaystyle \frac{\overline{\mathbf{F}}}{\overline{\mathbf{r}}}}d\overline{\mathbf{r}}+vp\int \underset{G}{\int }{\displaystyle \frac{{\mathbf{r}}_{0}div\overline{\mathbf{F}}+curl\overline{\mathbf{F}}\times {\mathbf{r}}_0}{r}}d\mathbf{S}]=-\overline{\mathbf{F}}\left(\mathbf{0}\right),$$

(69)

where $\overline{\mathbf{r}}grad\mathbf{F}=-\mathbf{F}$ and $\mathbf{r}div\overline{\mathbf{F}}-\mathbf{r}\times curl\overline{\mathbf{F}}=-\overline{\mathbf{F}}$. These vector integral formulas are analogous to the Cauchy-Pompeiu integral formula of complex analysis [17].

On the basis of the previous results one can say that there is a complete analogy between complex analysis in $\mathbb{C}$ and real vector analysis in ${\mathbf{V}}_2$, thus all the results of complex analysis are applicable to scalar and vector fields in ${\mathbf{V}}_2$ and vice versa. In doing so, z is formally replaced by $\mathbf{r}$, and the imaginary unit i, more precisely the ivector į, is replaced by the vector $\widehat{\mathbf{e}}$ and vice versa ($z\rightleftarrows \mathbf{r}$ and į$\rightleftarrows \widehat{\mathbf{e}}$). This conclusion can be even more obvious if a formally analogous method of deriving previously obtained vector identities is applied to the field of complex vectors ${\mathbf{V}}_{\mathbb{C}}$, which corresponds to the ivector field (field of complex numbers) $\mathbb{C}$, in the sense of the correspondence: $\mathbf{1}\fallingdotseq$$\mathbf{e}$ and į≒ $\widehat{\mathbf{e}}$, where the unit vector $\mathbf{e}$and the pseudo-unit vector $\widehat{\mathbf{e}}(\widehat{\mathbf{e}}·\widehat{\mathbf{e}}=-1$) form an orthogonal basis of the field of complex vectors ${\mathbf{V}}_{\mathbb{C}}$, whose algebraic structure is based on the geometric product of two complex vectors ${\mathbf{r}}_1=a\mathbf{e}+b\widehat{\mathbf{e}}\mathrm{and}{\mathbf{r}}_2=c\mathbf{e}+d\widehat{\mathbf{e}}$ as follows [9]

$${\overline{\mathbf{r}}}_1{\mathbf{r}}_2={\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2+{\mathbf{r}}_1{\overline{\mathbf{r}}}_2)+{\displaystyle \frac{1}{2}}({\overline{\mathbf{r}}}_1{\mathbf{r}}_2-{\mathbf{r}}_1{\overline{\mathbf{r}}}_2):=({\mathbf{r}}_1·{\overline{\mathbf{r}}}_2)\mathbf{e}-({\overline{\mathbf{r}}}_1\times {\overline{\mathbf{r}}}_2)\times \mathbf{e}=$$

(70)

$$=({\overline{\mathbf{r}}}_1·{\mathbf{r}}_2)\mathbf{e}+({\mathbf{r}}_1\times {\mathbf{r}}_2)\times \mathbf{e}=[(a\mathbf{e}-b\widehat{\mathbf{e}})·(c\mathbf{e}+d\widehat{\mathbf{e}})]\mathbf{e}+[(a\mathbf{e}+b\widehat{\mathbf{e}})\times (c\mathbf{e}+d\widehat{\mathbf{e}})]\times \mathbf{e}.$$

3. Algebraic Structure of the Field of Vectors ${\mathbf{V}}_3$

The Euclidean space ${\mathbb{R}}^3$ consists of three Euclidean spaces ${\mathbb{R}}^2$, which means that the field of vectors ${\mathbf{V}}_3$, isomorphic to it, consists of three fields of vectors ${\mathbf{V}}_2$, with base vectors ${\mathbf{e}}_i$ and ${\widehat{\mathbf{e}}}_i(i=1,2,3)$, such that ${\mathbf{e}}_1={\widehat{\mathbf{e}}}_3,{\mathbf{e}}_2={\widehat{\mathbf{e}}}_1$ and ${\mathbf{e}}_3={\widehat{\mathbf{e}}}_2$. Hence, the three plane vectors ${〈\mathbf{a}〉}_i={〈a\mathbf{e}+\widehat{a}\widehat{\mathbf{e}}〉}_i$, where $a_i=\mathbf{a}·{\mathbf{e}}_i$ and ${\widehat{a}}_i=\mathbf{a}·{\widehat{\mathbf{e}}}_i$, are component vectors of an arbitrary space vector $\mathbf{a}\in {\mathbf{V}}_3$, such that $2\mathbf{a}={〈\mathbf{a}〉}_i{\mathbf{e}}^i$. An index repeated as subscript and superscript in a product represents summation over the range of the index, by the Einstein summation convention.

The commutative geometric product of two space vectors $\overline{\mathbf{a}}\in {\mathbf{V}}_3$ and $\mathbf{b}\in {\mathbf{V}}_3$ is defined as the sum of the geometric products of the component vectors ${〈\overline{\mathbf{a}}〉}_i$ and ${〈\mathbf{b}〉}_i$, as follows

$$\overline{\mathbf{a}}\mathbf{b}=\mathbf{a}\circ \mathbf{b}+\mathbf{a}\wedge \mathbf{b}={\displaystyle \frac{{〈\overline{\mathbf{a}}\mathbf{b}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}=[{〈\mathbf{a}\circ \mathbf{b}〉}_i+{〈\mathbf{a}\wedge \mathbf{b}〉}_i]{\displaystyle \frac{{\mathbf{e}}^i}{\sqrt[2]{3}}},$$

(71)

where ${〈\overline{\mathbf{a}}〉}_i={〈a\mathbf{e}-\widehat{a}\widehat{\mathbf{e}}〉}_i$. Clearly, ${〈\overline{\mathbf{a}}〉}_i{\mathbf{e}}^i=\mathbf{0}$. If

$$\mathbf{a}\overline{\mathbf{b}}={\displaystyle \frac{{〈\mathbf{a}\overline{\mathbf{b}}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{〈\overline{\mathbf{b}}\mathbf{a}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}=[{〈\mathbf{b}\circ \mathbf{a}〉}_i+{〈\mathbf{b}\wedge \mathbf{a}〉}_i]{\displaystyle \frac{{\mathbf{e}}^i}{\sqrt[2]{3}}},$$

(72)

then $\mathbf{a}\circ \mathbf{b}=(\overline{\mathbf{a}}\mathbf{b}+\overline{\mathbf{b}}\mathbf{a})/2$ and $\mathbf{a}\wedge \mathbf{b}=(\overline{\mathbf{a}}\mathbf{b}-\overline{\mathbf{b}}\mathbf{a})/2$. In addition, the vector

$${\mathbf{a}}^{-1}={\displaystyle \frac{1}{\sqrt[2]{3}}}{\langle {\mathbf{a}}^{-1}\rangle }_i{\mathbf{e}}^i,$$

(73)

where ${\langle {\mathbf{a}}^{-1}\rangle }_i={\langle \overline{\mathbf{a}}/{∥\mathbf{a}∥}^2\rangle }_i$ and ${∥\mathbf{a}∥}_i^2={\left(a^2+{\widehat{a}}^2\right)}_i$, such that

$${\mathbf{a}}^{-1}\mathbf{a}={\displaystyle \frac{{〈{\mathbf{a}}^{-1}\mathbf{a}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{\mathbf{e}}_i{\mathbf{e}}^i}{\sqrt[2]{3}}},$$

(74)

is the inverse vector of the space vector $\mathbf{a}$, which allows division by the vector in ${\mathbf{V}}_3$. On the other hand, if $\langle a\widehat{b}-\widehat{a}b\rangle$ is denoted by the bracket $[a,\widehat{b}]$, it follows that

$$\mathbf{a}\wedge \mathbf{b}={\displaystyle \frac{{〈\mathbf{a}\wedge \mathbf{b}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{〈\mathbf{a}\times \mathbf{b}〉}_i\times {\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{[a,\widehat{b}]}_i{\widehat{\mathbf{e}}}^i}{\sqrt[2]{3}}}\mathrm{and}$$

(75)

$$\mathbf{a}\circ \mathbf{b}={\displaystyle \frac{{〈\mathbf{a}\circ \mathbf{b}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{〈\mathbf{a}·\mathbf{b}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}={\displaystyle \frac{{\langle ab+\widehat{a}\widehat{b}\rangle }_i{\mathbf{e}}^i}{\sqrt[2]{3}}}.$$

(76)

For a spatial vector field $\mathbf{F}={〈\mathbf{F}〉}_i{\mathbf{e}}^i={\langle F{\mathbf{r}}_0+F_{\perp }{\mathbf{r}}_{0\perp }\rangle }_i{\mathbf{e}}^i$, such that

$${〈\mathbf{F}d\overline{\mathbf{r}}〉}_i={〈(Fdr+r{F}_{\perp }d\phi )\mathbf{e}+(F_{\perp }dr-rFd\phi )\widehat{\mathbf{e}}〉}_i,$$

(77)

the geometric products

$${!}_{\overline{r}}=\mathbf{F}d\overline{\mathbf{r}}={\displaystyle \frac{{〈\mathbf{F}d\overline{\mathbf{r}}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}\mathrm{and}{\overline{!}}_r=\overline{\mathbf{F}}d\mathbf{r}={\displaystyle \frac{{〈\overline{\mathbf{F}}d\mathbf{r}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}},$$

(78)

where $2d\mathbf{r}={〈d\mathbf{r}〉}_i{\mathbf{e}}^i$

$$d{!}_{\overline{r}}=-2{ð}_{\mathbf{r}}\mathbf{F}d\widehat{\mathbf{S}}=-2{\displaystyle \frac{{\langle {ð}_{\mathbf{r}}\mathbf{F}d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i}{\sqrt[2]{3}}}\mathrm{and}$$

(79)

$$d{\overline{!}}_r=2\overline{{ð}_{\mathbf{r}}\mathbf{F}}d\widehat{\mathbf{S}}=2{\displaystyle \frac{{\langle \overline{{ð}_{\mathbf{r}}\mathbf{F}}d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i}{\sqrt[2]{3}}},$$

(80)

where $2d{\widehat{\mathbf{S}}}_i={\langle \mathbf{d}\overline{\mathbf{r}}\wedge \overline{\mathbf{d}}\mathbf{r}\rangle }_i$. According to (34)

$${\langle vt{\int }_{\partial S^{+}}^{⥀}\mathbf{F}\circ d\mathbf{r}\rangle }_i{\mathbf{e}}^i=-2{\langle vt\int \underset{G^{+}}{\int }({ð}_{\overline{\mathbf{r}}}\wedge \mathbf{F})d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i\mathrm{and}$$

(81)

$${\langle vt{\int }_{\partial S^{+}}^{⥀}\mathbf{F}\wedge d\mathbf{r}\rangle }_i{\mathbf{e}}^i=2{\langle vt\int \underset{G^{+}}{\int }({ð}_{\overline{\mathbf{r}}}\circ \mathbf{F})d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i,$$

(82)

so that

$$vt{\int }_{\partial S^{+}}^{⥀}{!}_{\overline{r}}=vt{\int }_{\partial S^{+}}^{⥀}\mathbf{F}d\overline{\mathbf{r}}={\displaystyle \frac{1}{\sqrt[2]{3}}}{\langle vt{\int }_{{\gamma }^{+}}^{⥀}\mathbf{F}d\overline{\mathbf{r}}\rangle }_i{\mathbf{e}}^i=$$

(83)

$$=-{\displaystyle \frac{2}{\sqrt[2]{3}}}{\langle vt\int \underset{G^{+}}{\int }{ð}_{\mathbf{r}}\mathbf{F}d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i=-2{\int }_{S^{+}}{ð}_{\mathbf{r}}\mathbf{F}d\widehat{\mathbf{S}}={\int }_{S^{+}}d{!}_{\overline{r}}\mathrm{and}$$

$$vt{\int }_{\partial S^{+}}^{⥀}{\overline{!}}_r=vt{\int }_{\partial S^{+}}^{⥀}\overline{\mathbf{F}}d\mathbf{r}={\displaystyle \frac{1}{\sqrt[2]{3}}}{\langle vt{\int }_{{\gamma }^{+}}^{⥀}\overline{\mathbf{F}}d\mathbf{r}\rangle }_i{\mathbf{e}}^i=$$

(84)

$$={\displaystyle \frac{2}{\sqrt[2]{3}}}{\langle vt\int \underset{G^{+}}{\int }\overline{{ð}_{\mathbf{r}}\mathbf{F}}d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i=2{\int }_{S^{+}}\overline{{ð}_{\mathbf{r}}\mathbf{F}}d\widehat{\mathbf{S}}={\int }_{S^{+}}d{\overline{!}}_r,$$

where the regions $G_i$, bounded by closed smooth Jordan curves ${\gamma }_i$, are the normal projections of the smooth surface S in ${\mathbf{V}}_3$ onto ${\mathbf{V}}_2^i$. Furthermore, since, see (17),

$${\langle (\overline{{ð}_{\mathbf{r}}\mathbf{F}}-{ð}_{\mathbf{r}}\mathbf{F})d\widehat{\mathbf{S}}\rangle }_i·{\mathbf{e}}^i={\langle (curl\mathbf{F}\times \mathbf{e})\circ d\widehat{\mathbf{S}}\rangle }_i·{\mathbf{e}}^i=$$

(85)

$$={〈curl\mathbf{F}〉}_i·{\langle \mathbf{e}\times d\widehat{\mathbf{S}}\rangle }^i={〈curl\mathbf{F}〉}^i·d{\mathbf{S}}_i=curl\mathbf{F}·d\mathbf{S}\mathrm{and}$$

$${\mathbf{e}}^i\times {\langle (\overline{{ð}_{\mathbf{r}}\mathbf{F}}+{ð}_{\mathbf{r}}\mathbf{F})d\widehat{\mathbf{S}}\rangle }_i={\mathbf{e}}^i\times {\langle (div\mathbf{Fe}\times d\widehat{\mathbf{S}})\times \mathbf{e}\rangle }_i={〈div\mathbf{Fe}〉}^i\times d{\widehat{\mathbf{S}}}_i,$$

(86)

the previous two vector identities leads to the generalized Stokes integral identity

$$vt{\int }_{\partial S^{+}}^{⥀}\mathbf{F}·d\mathbf{r}=2vt{\int }_{S^{+}}({ð}_{\overline{\mathbf{r}}}\wedge \mathbf{F})·d\mathbf{S},$$

(87)

where $\mathbf{F}·d\mathbf{r}={〈\mathbf{F}\circ d\mathbf{r}〉}_i·{\mathbf{e}}^i$ and $d\mathbf{S}={\mathbf{e}}^i\times d{\widehat{\mathbf{S}}}_i$, as well as to the vector integral identity

$$vt{\int }_{\partial S^{+}}^{⥀}\mathbf{F}\times d\mathbf{r}=2vt{\int }_{S^{+}}{〈{ð}_{\overline{\mathbf{r}}}\circ \mathbf{F}〉}^i\times d{\widehat{\mathbf{S}}}_i,$$

(88)

where $\mathbf{F}\times d\mathbf{r}={\mathbf{e}}^i\times {〈\mathbf{F}\wedge d\mathbf{r}〉}_i$.

On the other hand, for an arbitrary region V in the field ${\mathbf{V}}_3$field $\mathbf{F}=F\mathbf{e}+\widehat{F}\widehat{\mathbf{e}}$ satisfies the conditions of the divergence (Gauss-Ostrogradsky) theorem, then a procedure, similar to that for obtaining the integral identity (34), leads to the following integral identities

$$vt{\int }_{\partial V}^{⥀}{\mathbf{F}}_i({\mathbf{n}}^i·d\mathbf{S})=vt{\int }_{\partial V}^{⥀}{\langle \mathbf{F}d\mathbf{S}\rangle }_i{\mathbf{e}}^i=2vt{\int }_V{\langle {ð}_{\overline{\mathbf{r}}}\circ F\mathbf{e}+{ð}_{\overline{\mathbf{r}}}\wedge \widehat{F}\widehat{\mathbf{e}}\rangle }_i{\mathbf{e}}^{i}dV\mathrm{and}$$

(89)

$$vt{\int }_{\partial V}^{⥀}{\langle \mathbf{F}\times \mathbf{n}\rangle }_i({\mathbf{n}}^i·d\mathbf{S})=-vt{\int }_{\partial V}^{⥀}{\langle \mathbf{F}d\widehat{\mathbf{S}}\rangle }_i{\mathbf{e}}^i=2vt{\int }_V{\langle {ð}_{\overline{\mathbf{r}}}\circ \widehat{F}\widehat{\mathbf{e}}+{ð}_{\overline{\mathbf{r}}}\wedge F\mathbf{e}\rangle }_i{\mathbf{e}}^{i}dV,$$

(90)

where ${\mathbf{n}}_i={\langle \mathbf{e}\times \widehat{\mathbf{e}}\rangle }_i$ and $d{\mathbf{S}}_i={\langle \widehat{\mathbf{e}}\times (\mathbf{e}\times d\widehat{\mathbf{S}})\rangle }_i$. Therefore, since $d{\overline{\mathbf{S}}}_i={\langle d\mathbf{S}-d\widehat{\mathbf{S}}\rangle }_i$ it follows that

$$vt{\int }_{\partial V}^{⥀}\mathbf{F}d\overline{\mathbf{S}}=2vt{\int }_V{ð}_{\mathbf{r}}\mathbf{F}dV.$$

(91)

In addition, if $2{\langle {ð}_{\overline{\mathbf{r}}}\circ \mathbf{F}\rangle }_i·{\mathbf{e}}^i=div\mathbf{F}$ and ${\mathbf{e}}^i\times 2{\langle {ð}_{\overline{\mathbf{r}}}\wedge \mathbf{F}\rangle }_i=curl\mathbf{F}$, then

$$vt{\int }_{\partial V^{+}}^{⥀}\mathbf{F}·d\mathbf{S}=vt{\int }_{V^{+}}div\mathbf{F}dV\mathrm{and}$$

(92)

$$vt{\int }_{\partial V^{+}}^{⥀}d\mathbf{S}\times \mathbf{F}=vt{\int }_{V^{+}}curl\mathbf{F}dV,$$

(93)

which is a generalization of the well-known integral formulas in field theory.

In this acute case, $\mathbf{F}d\overline{\mathbf{S}}$ is the vector differential form ${!}_{\overline{S}}$, such that

$$d{!}_{\overline{S}}=2{ð}_{\mathbf{r}}\mathbf{F}dV=2{\displaystyle \frac{{〈{ð}_{\mathbf{r}}\mathbf{F}〉}_i{\mathbf{e}}^i}{\sqrt[2]{3}}}dV,$$

(94)

so that,

$$vt{\int }_{\partial V^{+}}^{⥀}{!}_{\overline{S}}=vt{\int }_{V^{+}}d{!}_{\overline{S}}.$$

(95)

4. Conclusions

Based on the integral identities obtained in the previous section, all fundamental integral identities, from Cauchy’s integral identity of complex analysis, through the Kelvin-Stokes (Green’s) integral identity and the Stokes integral identity, as well as Gauss-Ostrogradsky’s integral identity, all the way to the Newton-Leibniz formula [13], can be expressed by one vector integral identity

$$vt{\int }_{\partial {\Omega }^{+}}^{⥀}!=vt{\int }_{{\Omega }^{+}}d!,$$

(96)

where $\partial \Omega$ is the boundary of the corresponding compact region $\Omega$ in ${\mathbf{V}}_3$. For the Newton-Leibniz formula, the vector differential form $!$ is the vector field $\mathbf{F}=F\mathbf{e}$, such that

$$\Delta \mathbf{F}\left(I\right)=\left[F\left(b^{+}\right)-F\left(a^{-}\right)\right]\mathbf{e}=vt{\int }_{\partial I^{+}}^{⥀}!$$

(97)

is an interval vector field and $d!\left(\%\right)=2{ð}_{\mathbf{r}}\mathbf{F}\left(\%\right)d\mathbf{r}=\left[{lim}_{I\to \%}(\Delta \mathbf{F}/{\mu }_I)\right]d\mathbf{r}$, where I is some compact interval $\left[a,b\right]$ of the real line $\mathbb{R}$ and ${\text{-}}_I={\mu }_I\mathbf{e}$ is the vector Lebesgue measure of I. So, in the general case

$$2{ð}_{\mathbf{r}}\mathbf{F}\left(\%\right)=\underset{\Omega \to \%}{lim}{\displaystyle \frac{{\int }_{\partial \Omega }^{⥀}!}{{\text{-}}_{\Omega }}},$$

(98)

is the spatial derivative of the vector field $\mathbf{F}$ of the vector differential form $!$, and

$$Res\left(\mathbf{F},\%\right)={\displaystyle \frac{1}{2\pi \widehat{\mathbf{e}}}}\underset{\Omega \to \%}{lim}{\int }_{\partial \Omega }^{⥀}!$$

(99)

is the vector residue of $\mathbf{F}$ at the point $\%\in int.\Omega$.